Question

6. [2pt]
In the lab Springs, a student sits on a bosun's chair suspended by a spring. If the spring
has a spring constant, k = 1020 N/m, and is stretched by 0.5 m, how
much energy is stored in the spring?

Correct, computer gets: 127.5 J

7. [2pt]
The student gets off the bosun's chair very carelessly and all the energy in the
spring is converted into the kinetic energy of the seat. The mass of the seat is
1 kg. Once the spring has fully contracted, what is the speed of the seat?

Answer:  

8. [2pt]
Unfortunately, the student was hit on the head by the seat as it flew up. Some of the kinetic energy of the seat will be converted into
gravitational potential energy; neglect this. Assume the impact speed of
the seat is the same as in the previous question. If the seat comes to rest in 0.01 s as
it hits the student's head, what force would the seat exert on the student's
head?

Be safe. Remember: get off the bosun's chair carefully!

Answer:  

I GOT QUESTION 6, BUT NOT 7 AND 8.

Answer 1

6. The potential energy stored in a springis given by

                            

Where, k is spring constant and 'x' is expansion/compression in spring.

Given, k = 1020 N/m and x = 0.5 m

So,

                          

Hence, the potential energy stored in the spring is 127.5 J.

7. From the work energy theorem

                         Total initial energy = total final energy

Initially the energy stored in the chair (spring) = 127.5 J

When the student gets off from the chair, the total stored potential energy in the spring imediately gets converted into kinetic energy.

Therefore, new kinetic energy of the chair = 127.5 J

Kinetic energy is given by

                                  

Where, m is mass of the object , and v is the speed of the object

Given, mass (m) of the chair = 1 kg

therefore, the speed of the chair is

                             

Hence, the speed of the seat is 15.97 m/s.

8. According to the given information, the momentum of the chair changes to zero in just 0.01 second.

The momentum of the chair before the impact was

                                         mv = 1 x 15.97 = 15.97 kg.m/s .

So, change in momentum = Final momentum - initial momentum = 0 - 15.97 = -15.97 kg.m/s

So, applied force = (Change in momentum)/(time) = -15.97/0.01 = -1597 N.

Hence, the seat exerted a force of 1597 N on the head of the student's head.

For any doubt please comment.